What is the integral of (1/((1-sin(x)))) ?

The integral of (1/((1-sin(x)))) is -2/(tan(x/2)-1)+C

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Evaluate the Integral integral of 1/(1-sin(x)) with respect to x

\int (\frac{1}{( 1 - sin ( x ) )}) d x

- \frac{2}{tan ( \frac{x}{2} ) - 1} + C

Solution steps

\int \frac{1}{1 - sin ( x )} d x

Apply u-substitution

: \int \frac{2}{1 + u ^{2} - 2 u} d u

= \int \frac{2}{1 + u ^{2} - 2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{1 + u ^{2} - 2 u} d u

Complete the square

1 + u^{2} - 2 u : (u - 1)^{2}

= 2 \cdot \int \frac{1}{( u - 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 2 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 2 (- \frac{1}{v})

Substitute back

= 2 (- \frac{1}{tan ( \frac{x}{2} ) - 1})

Simplify

2 (- \frac{1}{tan ( \frac{x}{2} ) - 1}) : - \frac{2}{tan ( \frac{x}{2} ) - 1}

= - \frac{2}{tan ( \frac{x}{2} ) - 1}

Add a constant to the solution

= - \frac{2}{tan ( \frac{x}{2} ) - 1} + C

\frac{d}{d x} y = 8^{x}

\frac{d}{d x} (tan (\frac{1}{x}))

\frac{( cos ( 2 x ) )}{2}

(- 4, 4), (4, 4)

\frac{d}{d x} (\frac{x}{5})

What is the integral of (1/((1-sin(x)))) ?
The integral of (1/((1-sin(x)))) is -2/(tan(x/2)-1)+C